Question

a ball is thrown vertically upward with the velocity of 10 m/s calculate hieght ataind by the ball time to the reach heighest if the acclerattion of -10m/s​

Answer 1

Explanation:

v^2=u^2+2as

0=10^2+(2×-10×s)

0=100-20s

20s=100

s=100/20

s=5m answer

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Answer 2

$\sf\large{\underline{\underline{Given:}}}$

$\sf{\longrightarrow Innitial\:velocity\:(u) = 10\:m{s}^{-1}}$

$\sf{\longrightarrow Acceleration\:(a) = -10\:m{s}^{-1}}$

$\sf{then,\: \longrightarrow final \:velocity\:(v) = 0\:m{s}^{-1}}$

$\sf\large{\underline{\underline{To\:Find:}}}$

$\sf{\longrightarrow Height\:attended\:by\:the\:ball\:(h) = \: \: ...?}$

$\sf\large{\underline{\underline{SolUTIoN:}}}$

$\sf\red{\underline{\underline{Using\: ({v}^{2} = {u}^{2} + 2as)}}}$

$\sf{\implies ({0}^{2}) = ({10}^{2}) + 2 \times -10 \times s}$

$\tt{\implies 0 = 100 + (-20\:s)}$

$\tt{\implies -100 = -20\:s}$

$\tt{\implies (\dfrac{-100}{-20}) = s}$

$\tt{\implies 5 = s}$

$\sf\large{\underline{\underline{Hence:}}}$

$\sf{\longrightarrow Height\:attended\:by\:the\: ball = 5\:m}$